The animation above (rotateable!) shows the motion of the four birds. From there, it is obvious that the birds will not stay in a tetrahedronal shape. Fortunately, there is a perspective from which the birds will stay in a consistent configuration: when viewed from the axis (1, 1, 0), the birds are configured in a square. Also, the velocities of the 4 birds possess a rotational symmetry about this axis, thanks to which the birds will always remain in a square when viewed from the axis.

It might seem that this problem now reduces down to the well-known problem of four turtles/bugs in a square. However, this is not the case as the birds also fly in the direction of the axis.

Let us label the distance of bird A from the centroid along the axis as \(h\). This distance is therefore \(-h\) for bird B, \(h\) for C and \(-h\) for D. All birds are at the radius r from the axis and consecutive birds form an angle \(\frac{\pi}{2}\) with the axis.

Consecutive birds, say A and B, are separated by a distance $$| \vec{AB}| =\sqrt{2r^2+4h^2}$$ between them. The magnitude of bird A's velocity along the axis is $$\frac{dh}{dt}=-\frac{2h}{\sqrt{2r^2+4h^2}}v$$ and it's radial velocity is $$\frac{dr}{dt}=-\frac{r}{\sqrt{2r^2+4h^2}}v.$$

Putting these expressions together, we get $$\frac{dh}{dr}=\frac{2h}{r} \quad \text{ or } \quad \ln h = \ln r + C.$$ Knowing that initially \(r_0=\frac{1}{2}\) and \(h_0=\frac{\sqrt{2}}{4}\), we get $$h=\sqrt{2}r^2.$$

Knowing this is enough to find the distance which the bird flies before the collisions. In a square configuration, the radial velocity of any bird is always equal to its tangential velocity, that is why in a time interval \(dt\) the bird will traverse a path \(\sqrt{2(dr)^2+(dh)^2}=\sqrt{2}\sqrt{1+2r^2}\,dr\). Hence, the total traversed path is $$\int_0^{1/2} \sqrt{2}\sqrt{1+4r^2}\,dr= \frac{\text{arcsinh}(1)+\sqrt{2}}{2^{\frac{3}{2}}}\approx 0.812$$